Problem: Simplify the following expression and state the condition under which the simplification is valid: $z = \dfrac{t^2 + 6t - 16}{t^2 + 8t}$
Solution: First factor the expressions in the numerator and denominator. $ \dfrac{t^2 + 6t - 16}{t^2 + 8t} = \dfrac{(t - 2)(t + 8)}{(t)(t + 8)} $ Notice that the term $(t + 8)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(t + 8)$ gives: $z = \dfrac{t - 2}{t}$ Since we divided by $(t + 8)$, $t \neq -8$. $z = \dfrac{t - 2}{t}; \space t \neq -8$